# Linear System with Parameter

Here is problem 3 from the Fifth International Internet Mathematical Olympiad for Students. This is an online competition run by the Ariel University Center of Samaria, Israel.

Find all values of the parameter y for which the following system has a solution

 (1) x5 + x2 = yx1 (2) x1 + x3 = yx2 (3) x2 + x4 = yx3 (4) x3 + x5 = yx4 (5) x4 + x1 = yx5

For each value of y, find all the solutions.

Solution

Find all values of the parameter y for which the following system has a solution

 (1) x5 + x2 = yx1 (2) x1 + x3 = yx2 (3) x2 + x4 = yx3 (4) x3 + x5 = yx4 (5) x4 + x1 = yx5

For each value of y, find all the solutions.

First of all observe that, for any value of y, the system has a trivial solution: x1 = x2 = x3 = x4 = x5 = 0. The task is to find the values of y for which there is a nontrivial solution. Add up all five equations to obtain

 (y - 2)(x1 + x2 + x3 + x4 + x5) = 0.

So, either y = 2 or x1 + x2 + x3 + x4 + x5 = 0. Clearly the value y = 2 is bound to play a prominent role in the solution. Curiously, there are other y's of no lesser importance.

But, to follow up, assume first that y = 2. The system becomes

 (1') x5 + x2 = 2x1 (2') x1 + x3 = 2x2 (3') x2 + x4 = 2x3 (4') x3 + x5 = 2x4 (5') x4 + x1 = 2x5

Since the system is homogeneous and we are looking for a non-trivial solution, we may omit the last equation and solve the rest with the Gaussian elimination, getting subsequently,

 - 2 1 0 0 1 1 - 2 1 0 0 0 1 - 2 1 0 0 0 1 - 2 1
⇒
 - 2 1 0 0 1 0 - 3 2 0 1 0 0 - 4 3 1 0 0 0 - 5 5

By the back substitution, x5 = x4 = x3 = x2 = x1, with arbitrary x1.

Now we may assume y ≠ 2 and x1 + x2 + x3 + x4 + x5 = 0. We'll replace (quite arbitrarily) the last equation with this one and employ again the Gaussian elimination.

 1 1 1 1 1 -y 1 0 0 1 1 -y 1 0 0 0 1 -y 1 0 0 0 1 -y 1
⇒
 1 1 1 1 1 0 y+1 y y y+1 0 -(y+1) 0 -1 -1 0 1 -y 1 0 0 0 1 -y 1
⇒

 1 1 1 1 1 0 1 -y 1 0 0 y+1 y y y+1 0 -(y+1) 0 -1 -1 0 0 1 -y 1
⇒
 1 1 1 1 1 0 1 -y 1 0 0 0 y y-1 y 0 0 -y(y+1) y -1 0 0 1 -y 1
⇒

 1 1 1 1 1 0 1 -y 1 0 0 0 -y(y+1) y -1 0 0 y y-1 y 0 0 y+1 -1 y+1
⇒
 1 1 1 1 1 0 1 -y 1 0 0 0 -y(y+1) y -1 0 0 0 y²+y-1 y²+y-1 0 0 0 0 y²+y-1

If y² + y - 1 ≠ 0, the last equation gives x5 = 0 and the back substitution leads to a trivial solution which was already accounted for. So let us assume that y² + y - 1 = 0. The quadratic formula gives two distinct solutions

 y1,2 = (-1 ± √5) / 2.

With the parameter y equal to one of them, the last equation implies that x5 can be chosen arbitrarily; and, from the fourth equation, the same is true of x4.

Since, for the chosen y, -y(y + 1) = -1, the third equation gives

 x3 = yx4 - x5

From the second equation,

 x2 = -y(x4 + x5)

and, from the first equation,

 x1 = -x4 + yx5.

Thus the system has three sets of solutions:

1. A trivial solution for any y,

2. A 1-parameter solution for y = 2,

3. A 2-parameter solution for y = (-1 ± 5) / 2.

The latter family, perhaps surprisingly, includes the golden ratio.